y = x − a sin ⁡ ( b x ) − 3where x = x ( a , b )We have to use implicit differentiation to calculate d x d a , d x d b .This is what I get. d y d a = d x d a − sin ⁡ ( b x )for d x d a and d y d b = d x d b − a b cos ⁡ ( b x ) d x d b for d x d b Can someone please let me know if I am doing it right. Thanks

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y  =  x  −  a  sin  ⁡  (  b  x  )  −  3where   x  =  x  (  a  ,  b  )We have to use implicit differentiation to calculate             d      x              d      a        ,            d      x              d      b      .This is what I get.            d      y              d      a        =            d      x              d      a        −  sin  ⁡  (  b  x  )for             d      x              d      a       and            d      y              d      b        =            d      x              d      b        −  a  b  cos  ⁡  (  b  x  )            d      x              d      b      for             d      x              d      b      Can someone please let me know if I am doing it right. Thanks

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Write your function:  y  =  x  (  a  ,  b  )  −  a  sin  ⁡      (    b    x    (    a    ,    b    )    )    −  3to remeber that   x is a function of   a and   bNow you have:                    ∂        y                    ∂        a              =                    ∂        x        (        a        ,        b        )                    ∂        a              −            ∂              ∂        a                  [    a    sin    ⁡          (      b      x      (      a      ,      b      )      )        ]  using the product rule on the second term you find:                    ∂        y                    ∂        a              =                    ∂        x                    ∂        a              −      [                            ∂          a                          ∂          a                      sin    ⁡          (      b      x      (      a      ,      b      )      )        +    a                  ∂                  ∂          a                      sin    ⁡          (      b      x      (      a      ,      b      )      )        ]  Now using the chain rule:                    ∂        y                    ∂        a              =                    ∂        x                    ∂        a              −  sin  ⁡      (    b    x    )    −  a  b  cos  ⁡      (    b    x    )                      ∂        x                    ∂        a            In the same manner you can find                     ∂        y                    ∂        b            EDIT: but rereading i&#039;m not sure this is the question. The OP is a bit confused about   x and   y

y=x−asin(bx)−3 where x=x(a,b) We have to use implicit differentiation to calculate (dx)/(da),(dx)/(db).

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